Question

How do you find the roots, real and imaginary, of `y= -3x^2 -2x +(2x+1)^2` using the quadratic formula?

Answer 1

`y=-3x^2-2x+(2x+1)^2` has a double root at `x=-1`

Explanation:

`y=-3x^2-2x+(2x+1)^2`
`y=-3x^2-2x+(4x^2+4x+1)`
`y=-3x^2-2x+4x^2+4x+1`
`y=x^2+2x+1`

Using the quadratic formula
`x=(-2+-sqrt(4-4(1)(1)))/(2)`
`x=(-2+-sqrt0)/2`
`x=(-2)/2`
`x=-1`

Therefore, `y=-3x^2-2x+(2x+1)^2` has a double root at `x=-1`